Exercise 26.37 of PSLS 3e ------------------------- Data on the resistance of corn hybrids (produced by genetic engineering) to a certain herbicide, glusofinate. In this study, resistance is indicated by low degrees (percents) of leaf burn (necrosis) after application of the herbicide. Three corn hybrids (two resistant and one non-resistant) were used. Six different application rates (doses) of the herbicide were used, ranging from zero (control) to 1.44 kg/ha. The dataset has four replications per combination of application rate and hybrid; it is not described what these are or what the observational unit is (e.g. plant or plot). (a) Minitab commands for a two-way table of descriptive statistics as well as an ANOVA analysis including the requested plot of means (interaction plot) as well as residual plots: MTB > WOpen "R:\Chapter 26\ex26_037.mtw". Retrieving worksheet from file: ‘R:\Chapter 26\ex26_037.mtw’ Worksheet was saved on 22/11/2014 MTB > Table 'Rate' 'Hybrid'; SUBC> Layout 1 1; SUBC> DMissing 'Rate' 'Hybrid'; SUBC> Means 'Percent'; SUBC> StDev 'Percent'; SUBC> Counts. Tabulated statistics: Rate, Hybrid Rows: Rate Columns: Hybrid NonResistant Resistant1 Resistant2 All 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4 4 4 12 0.10 27.50 0.00 2.50 10.00 5.00 0.00 5.00 13.48 4 4 4 12 0.20 25.00 0.00 17.50 14.17 10.00 0.00 12.58 13.79 4 4 4 12 0.36 35.00 0.00 30.00 21.67 17.32 0.00 8.16 18.99 4 4 4 12 0.80 45.00 0.00 25.00 23.33 12.91 0.00 5.77 20.60 4 4 4 12 1.44 94.75 5.00 35.00 44.92 3.69 10.00 12.91 39.93 4 4 4 12 All 37.88 0.83 18.33 19.01 30.80 4.08 15.51 24.96 24 24 24 72 Cell Contents: Percent : Mean Percent : Standard deviation Count MTB > ANOVA 'Percent' = Rate Hybrid Rate* Hybrid; SUBC> GFourpack. ANOVA: Percent versus Rate, Hybrid Factor Type Levels Values Rate fixed 6 0.00, 0.10, 0.20, 0.36, 0.80, 1.44 Hybrid fixed 3 NonResistant, Resistant1, Resistant2 Analysis of Variance for Percent Source DF SS MS F P Rate 5 13955.1 2791.0 43.49 0.000 Hybrid 2 16481.7 8240.8 128.40 0.000 Rate*Hybrid 10 10318.5 1031.8 16.08 0.000 Error 54 3465.7 64.2 Total 71 44221.0 S = 8.01128 R-Sq = 92.16% R-Sq(adj) = 89.70% Residual Plots for Percent MTB > Interact 'Rate' 'Hybrid'; SUBC> Response 'Percent'; SUBC> Full. Interaction Plot for Percent Comments: --------- The interaction plot with rate on the x-axis and curves for each hybrid shows 3 clearly non-parallel curves. One of these (for hybrid Resistant1) is essentially flat at zero, and inspection of the data shows that all values are zero except for a single value of 20 at rate 1.44. The two other curves show increasing values with rate, both of starting at zero for rate zero (which is logical: without any application there should not be any leaf burn). The curve for the non- resistant hybrid has larger means across all non-zero rates, with the largest differences at rates 0.10 and 1.44. In this case, there does not seem any value in further describing the main effects because all the relevant information can be read off the interaction plot. (b) The two-way ANOVA model has the assumptions: - normal distribution of errors - same std.dev. (or variance) of errors across the data - independence of errors (or observations) - the same mean for the 4 observations within each combination of hybrid and rate. The four-in-one residual panel shows several unusual features. First, the normal probability plot has a quite discrete character, with many replicates of residuals with the same value. In particular, there seems to be many residuals equal to zero. The histogram has a very strong peak at zero and the plot against order shows a block of zeros between 20 and 40. It is clear that a normal distribution does not describe the residuals well. Also the plots of stand. residuals against fitted values shows discrete patterns. The observation order here is determined by the factor and does therefore not have any additional meaning, but it is clear that very many residuals are equal to zero (in fact, 30 out 72 values). The reason for this is that 7 out of the 18 groups have only zero percents. The leads to the main issues with the model assumptions: the data are too discrete, and the variance is not at all constant across the data because large parts of the data show no variation (in particular, the corn hybrid Resistant1 has only one non-zero observation). Therefore the two-way ANOVA model does not meet its assumptions, and the validity of the analysis is uncertain (to say the least). One possible option for analysis of the data is to average the replicates (whose meaning is not clear from the description provided anyway) to a single value per cell, and then use Friedman's test. That could give an overall comparison of the 3 corn hybrids, but the interaction cannot be explored after averaging, so this does not seem very attractive. Another option is to restrict analysis to the two other hybrids than Resistant1 (which anyway shows clearly different results, for which no statistical assessment seems needed) and non-zero rates. For demonstration purposes, Minitab code for the suggested analyses, without any further comments: MTB > Subset; SUBC> Exclude; SUBC> Equals Hybrid "Resistant1"; SUBC> Name "Hybrid equals Resistant1 excluded". Subset worksheet Hybrid equals Resistant1 excluded created. Results for: Hybrid equals Resistant1 excluded MTB > ANOVA 'Percent' = Rate Hybrid Rate* Hybrid; SUBC> GFourpack. ANOVA: Percent versus Rate, Hybrid Factor Type Levels Values Rate fixed 6 0.00, 0.10, 0.20, 0.36, 0.80, 1.44 Hybrid fixed 2 NonResistant, Resistant2 Analysis of Variance for Percent Source DF SS MS F P Rate 5 19420.1 3884.0 44.17 0.000 Hybrid 1 4582.5 4582.5 52.11 0.000 Rate*Hybrid 5 4770.1 954.0 10.85 0.000 Error 36 3165.7 87.9 Total 47 31938.5 S = 9.37750 R-Sq = 90.09% R-Sq(adj) = 87.06% Residual Plots for Percent Results for: ex26_037.mtw MTB > Name c4 "ByVar1" c5 "ByVar2" c6 "Mean1" MTB > Statistics 'Percent'; SUBC> By 'Rate' 'Hybrid'; SUBC> GValues 'ByVar1'-'ByVar2'; SUBC> Mean 'Mean1'. MTB > name 'ByVar1' 'Rate2' MTB > name 'ByVar2' 'Hybrid2' MTB > name 'Mean1' 'meanPercent' MTB > Friedman 'meanPercent' 'Rate2' 'Hybrid2'. Friedman Test: meanPercent versus Rate2 blocked by Hybrid2 S = 9.86 DF = 5 P = 0.079 S = 12.18 DF = 5 P = 0.032 (adjusted for ties) Est Sum of Rate2 N Median Ranks 0.00 3 0.00 5.0 0.10 3 9.17 8.0 0.20 3 15.42 8.0 0.36 3 25.42 12.0 0.80 3 25.00 12.0 1.44 3 35.00 18.0 Grand median = 18.33